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    Metacentre & Metacentric Height of A Ship (02042020)

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    Xahid Hasan
    The document discusses metacentre and metacentric height of ships. It defines metacentre as the point of intersection between the center of buoyancy lines of an upright and inclined ship. Metacentric height is the distance between the metacentre and center of gravity, with larger heights indicating greater stability. It then derives the expression for calculating metacentric height (KMT) and metacentric radius (BM), showing that BM=I/Δ, where I is the second moment of area and Δ is the displacement volume.

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    Metacentre & Metacentric Height of A Ship (02042020)

    Uploaded by

    Xahid Hasan
    176 views2 pages
    The document discusses metacentre and metacentric height of ships. It defines metacentre as the point of intersection between the center of buoyancy lines of an upright and inclined ship. Metacentric height is the distance between the metacentre and center of gravity, with larger heights indicating greater stability. It then derives the expression for calculating metacentric height (KMT) and metacentric radius (BM), showing that BM=I/Δ, where I is the second moment of area and Δ is the displacement volume.

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    5. Metacentre & Metacentric Height of a Ship (02042020)

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    The document discusses metacentre and metacentric height of ships. It defines metacentre as the point of intersection between the center of buoyancy lines of an upright and inclined ship. Metacentric height is the distance between the metacentre and center of gravity, with larger heights indicating greater stability. It then derives the expression for calculating metacentric height (KMT) and metacentric radius (BM), showing that BM=I/Δ, where I is the second moment of area and Δ is the displacement volume.

    Copyright:

    © All Rights Reserved
    Download as DOCX, PDF, TXT or read online from Scribd
    Download as docx, pdf, or txt
    176 views2 pages

    Metacentre & Metacentric Height of A Ship (02042020)

    Uploaded by

    Xahid Hasan
    The document discusses metacentre and metacentric height of ships. It defines metacentre as the point of intersection between the center of buoyancy lines of an upright and inclined ship. Metacentric height is the distance between the metacentre and center of gravity, with larger heights indicating greater stability. It then derives the expression for calculating metacentric height (KMT) and metacentric radius (BM), showing that BM=I/Δ, where I is the second moment of area and Δ is the displacement volume.

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    © All Rights Reserved
    Download as DOCX, PDF, TXT or read online from Scribd
    Download as docx, pdf, or txt
    of 2

    Discuss metacentre and metacentric height of a ship?

    Derive the expressions


    for finding the position of metacentre of a ship or Derive the expression
    BM=I/▼, Where symbols have their usual meaning.

    Metacentre:
    Metacenter is the point of intersection between two vertical lines, one line through the
    center of buoyancy of the hull of a ship in equilibrium and the other line through the
    center of buoyancy of the hull when the ship is inclined to one side; the distance of
    this intersection above the center of gravity is an indication of the stability of the ship.
    In the figure below MT is the metacentre of a ship.

    Metacentric Height:
    Metacentric height is the distance between a floating body’s metacenter and its center
    of gravity. It is a measure of a floating body’s stability such that a ship with a large
    metacentric height is more stable than a ship with a smaller one. A ship that has a
    large metacentric height also has a shorter rolling. In the figure below GMT is the
    metacentric height of a ship. On the other hand KMT is the metacentric height above
    the keel of a ship. From the figure, KMT = KB + BMT, where KB is the vertical
    distance of centre of buoyancy above the keel.

    Expressions for finding the position of metacentre of a ship:


    The distance of the transverse metacentre above the keel (KMT) is given by KMT = KB
    + BMT.

    KB is the distance of the centre of buoyancy above the keel and may be found by one
    of the methods shown previously.
    y
    BM may be found as follows:
    Consider a ship whose volume of displacement is ▼, lying upright at waterline WL,
    the centre of buoyancy being on the centreline of the ship. If the ship is now inclined
    to a small angle Φ, it will lie at waterline WLl which intersects the original waterline
    at S (Fig). Since Φ is small it may be assumed that S is on the centreline.
    A wedge of buoyancy WSWI has been moved across the ship to LSL1 causing the
    centre of buoyancy to move from B to B1

    Let v = volume of wedge


    gg1 = transverse shift in centre of gravity of wedge
    Then BB1=v x gg1/▼
    But BBI =BM tan Φ

    So BM tan Φ= v x gg1/▼
    or, BM= v x gg1/(▼x tan Φ)

    To determine the value of v x ggl, divide the ship into thin transverse strips of length
    dx, and let the half width of waterplane in way of one such strip be y=Breadth/2=B/2
    = Half Breadth of Ship.

    Volume of strip of wedge, v = 1/2 y x y tan Φ dx= 1/2 y2 tan Φ dx

    Mass of strip of wedge, m = p x v = p x 1/2 y2 tan Φ dx

    Moment of transfer of strip of wedge= m x gg1=4/3 y x p x 1/2 y2 tan Φ dx= p x 2/3 x y3 tan Φ dx

    Total moment of transfer of wedge= m x ggl = p tan Φ Sum 2/3 x y3 dx

    But Sum 2/3 x y3 dx = 2nd moment of area of free surface about the centreline of the tank=I
    Therefore, v x gg1 = I tan Φ

    and BM= v x gg1/(▼x tan Φ)


    = I tan Φ/(▼x tan Φ)
    BM= I/▼

    So BM= I/▼, which is the required expression. (Derived)

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